Problem: Solve for $x$ and $y$ using substitution. ${6x+5y = 1}$ ${x = -3y-2}$
Since $x$ has already been solved for, substitute $-3y-2$ for $x$ in the first equation. ${6}{(-3y-2)}{+ 5y = 1}$ Simplify and solve for $y$ $-18y-12 + 5y = 1$ $-13y-12 = 1$ $-13y-12{+12} = 1{+12}$ $-13y = 13$ $\dfrac{-13y}{{-13}} = \dfrac{13}{{-13}}$ ${y = -1}$ Now that you know ${y = -1}$ , plug it back into $\thinspace {x = -3y-2}\thinspace$ to find $x$ ${x = -3}{(-1)}{ - 2}$ $x = 3 - 2$ ${x = 1}$ You can also plug ${y = -1}$ into $\thinspace {6x+5y = 1}\thinspace$ and get the same answer for $x$ : ${6x + 5}{(-1)}{= 1}$ ${x = 1}$